Talk:Cantor's first set theory article/Archive 1

initial section

Is it really true that most mathematicians believe that the diagonal proof was Cantor's first proof of uncountability? I'm no mathematician, but even my topical interest in the matter turned up that fact long before this article was created. Is the misconception really that prevalent? -- Cyan 20:51, 3 Nov 2003 (UTC)

I haven't carefully polled mathematicians to ascertain this, but I keep finding it asserted in print, and I've spoken with a number of intelligent mathematicians who were under that impression. Michael Hardy 19:53, 4 Nov 2003 (UTC)

I would not have been certain if the diagonal proof was the first one, but my guess (if I would have to bet) would have been that it was, as this is the proof that is most known and famous, so in this sense, I think it's a misconception. Also, mathematicians are pretty sloppy historians (see Fermat number re: Gauss n-gon construction) so it's best to assume we don't know what we're doing, I think. Revolver

I've looked up Cantor's 1874 paper in Journal für die Reine und Angewandta Mathematik, and the argument given in that article is indeed the one given here. See also Joseph Dauben's book about Cantor. This was indeed his first proof of this result. Michael Hardy 22:17, 12 Jan 2004 (UTC)

Was it really in 1877 that Cantor discovered the diagonal method, or was it later? I cannot find any proof for this. -- Zwaardmeester 19:52, 15 Jan 2006 (GMT+1)

This "proof" is logically flawed

The constructivists' counterargument to the preceding "proof" of the "uncountability" of the set of all real numbers hinges on their firm belief that there is no greatest natural number --- hence, there is no last term in the sequence {Xn}. The completed constructibility of the monotone sequences {An} and {Bn} from {Xn} is also dubious for the same reason --- hence, the limit point C is "unreachable" (that is, it perpetually belongs to the elements of R denoted by the 3-dot ellipsis ". . ."). One could state in more familiar terminology that C is an irrational number which does not have a last, say, decimal expansion digit since there is no end in the progression of natural numbers.

For a simple counterargument, even granting arguendo Georg Cantor's own hierarchy of transfinite ordinal numbers, we merely note that the infinite set whose size is "measured" by the ordinal number, say, w+1 = {0,1,2,3,...,w} (here, w is omega)) is also countable. To elaborate, "ordinal numbers" are "order types of well-ordered sets". A well-ordering is an imposition of order on a non-empty set which specifies a first element, an immediate successor for every "non-last element", and an immediate successor for every non-empty subset that does not include the "last element" (if there is one). For examples: (1) The standard imposition of order on all the non-negative rational numbers: { 0, . . ., 1/4, . . ., 1/2, . . ., 3/4, . . ., 1, . . ., 5/4, . . ., 3/2, . . ., 7/4, . . ., 2, . . ., 9/4, . . . } is not a well-ordering — because, for example, there is no positive rational number immediately following 0. (2) The following imposition of order on all the non-negative rational numbers is a well-ordering but not a countable ordering or enumeration: { 0, 1, 2, 3, . . ., 1/2, 3/2, 5/2, . . ., 1/3, 2/3, 4/3, . . ., 1/4, 3/4, 5/4, . . ., 1/5, 2/5, 3/5, . . . } (3) The following imposition of order on all the non-negative rational numbers is a well-ordering that is also a countable ordering or enumeration: { 0, 1, 1/2, 2, 1/3, 3, 1/4, 2/3, 3/2, 4, 1/5, 5, 1/6, 2/5, 3/4, 4/3, 5/2, 6, 1/7, 3/5, 5/3, 7, . . . } It must be emphasized that, while the first two representations are not countable ordering or enumeration of the non-negative rational numbers, nevertheless, the set of non-negative rational numbers is countable as the third representation shows.

Rigorously, let us grant arguendo Georg Cantor's claim of completed totality of infinite sets. The sequences {An} and {Bn} were defined to form a sequence of nested closed intervals [An,Bn] that "microscopes" to the limit point C which must be a member of R. The assumption that all the elements of R can be enumerated in the sequence {Xn} carries with it the imposition of a countable ordering --- which deviates from the standard ordering based on the numerical values --- of the members of R. Given an arbitrary countably ordered sequence {Xn}, by the very definition of the construction of {An} and {Bn} stipulated by Georg Cantor, there is only one sequence of nested closed intervals [An,Bn] that can be so constructed --- hence, only one same limit point C for both {An} and {Bn}. In other words, the enumeration scheme of the sequence {Xn} determines exactly the sequences {An} and {Bn} as well as the limit point C. A rearrangement of a finite number of terms of {Xn} is still a countable ordering of the elements of R --- if a rearrangement results in different sequences {An} and {Bn}, then a different limit point C is obtained but, always, there is only one limit point C that is "excluded" in any specified sequence {Xn}. We emphasize that in Cantor's tenet of completed totality of an infinite set, the limit point C of both the sequences {An} and {Bn} is known "all at once" in advance.

Therefore, Georg Cantor could have just as well specified the also countable set {X1,X2,X3,...} U {C} = {X1,X2,X3,...,C} [note that there is no real number immediately preceding C] --- instead of the standard enumeration {X1,X2,X3,...} that he assumed in the first sentence of his "proof" as having all of R as its range so that C is clearly not in the sequence {X1,X2,X3,...} but C is in R = {X1,X2,X3,...,C} and no contradiction would be reached.

Furthermore, {Xn} = {Yn} U {An} U {Bn} U {C} [1] where the sequence {Yn} consists of all the terms Xi of the sequence {Xn} that were bypassed in the construction of the sequences {An} and {Bn}. If we accept Cantor's reductio ad absurdum argument above, then we deny equality [1] and the easily proved fact that the union set of a finite collection of countable sets is countable.

BenCawaling@Yahoo.com

After carefully reading the above post, I conclude that the author is making a mistake similar to those made by many people encountering diagonal-method proofs for the first time. The problem is that if the initial set chosen is {X1,X2,X3, ... C} (with C inserted between two Xi's -- this is not captured by the notation), then the limit value produced by Cantor's argument will not be C, but something else.

Even a good high school student would balk at the above objection --- are you saying that a monotone sequence could have a middle-term limit? [BenCawaling@Yahoo.com (27 Sep 2005)]

No. You're saying "For a given proposed enumeration (x_n) of the reals, Cantor presents a number c which is not captured. So let's just throw c in and we have enumerated the reals." The objection to your argument is: once you have thrown in c, Cantor will happily present a new number (different from c) that is still not covered by your new proposed enumeration. He will always catch you. AxelBoldt 18:17, 30 March 2006 (UTC)

This is the same commonest mistake in Cantor's diagonal argument --- once you have thrown in c and claim a complete list of the real numbers then you're done; a re-application of Cantor's diagonal argument does present a number (different from c) but that new anti-diagonal number was included in the list when c was the anti-diagonal number. You merely made a new enumeration or re-arranged the row-listed real numbers to get a different anti-diagonal number! → I hope you teach my counter-counterargument to your counterargument in your math classes ... Best regards ... [BenCawaling@Yahoo.com (16 Oct 2006)]

"...but that new anti-diagonal number was included in the list when c was the anti-diagonal number." No it wasn't! Ossi 19:47, 8 September 2007 (UTC)

It is wise to understand that any real number is a constant [that is, in place-value positional numeral system like decimal or binary, its every digit is known (or could be computed “given sufficient computing time and space resources” which is a standard axiom in the modern theory of computation)] so they have fixed existence (unlike a variable) and there could not be any algorithm such as Cantor’s anti-diagonalization argument that can produce “new” numbers (that is, not already on the row-list unless it is an extended number such as row-listing all the rationals and having irrational diagonal and anti-diagonal number or row-listing all the real numbers and having complex diagonal and anti-diagonal number) like the 2nd anti-diagonal number, say D, when the first anti-diagonal number C is in the row-list. Any real number is either the anti-diagonal number or in the “actually incomplete” row-listing. If D was not included in the row-listing when C was the anti-diagonal number, then you don’t have to call upon the excluded anti-diagonal number C because your first row-list is already incomplete since it also excludes another real number D. In the sequence of ALL real numbers {X_n} = {Y_n} U {A_n} U {B_n} U {C} of my discussion above, C is an arbitrary real number limit. If the 2nd limit D is not equal to the first limit C, then D is in {Y_n} or in {A_n} or in {B_n} when C is the limit, and vice versa on “C” and “D” — otherwise, {X_n} does not enumerate all the real numbers as presupposed.

From a different perspective, the following are tautologous diagonalization (not anti-diagonalization like Cantor’s) arguments:

* a truly complete row-listing with 1st row the positive integers in their respective columns; in row 2, the respective positive integers 2nd powers (squares) ; in row 3, the respective positive integers 3rd powers (cubes); …; in row n, the respective positive integers nth powers; … . Then, clearly, the diagonal sequence of nⁿ is not in the row-list (Leo Zippin in “Uses of Infinity” [MAA, 1962] claims this to be an example of Cantor’s diagonal argument);

* a truly complete enumeration or row-listing of all the fractional rational numbers would have irrational diagonal and anti-diagonal numbers;

* a truly complete enumeration or row-listing of all the fractional real numbers — which consists of all the rational (nonterminating but periodic binary or decimal expansion) and irrational (nonterminating and nonperiodic expansion) fractional numbers with each collection exhausting the positive integer row-indices and the representation 0.b₁b₂b₃… (for the rational numbers, there is no limit in the period-length while for the irrational numbers, each has a nonzero digit at the omegath position due to the 1-1 correspondence, say π-3 in binary, <0.0, 0.00, 0.001, 0.0010, 0.00100, 0.001001, …, π-3> <--> <0, 0, 1, 0, 0, 1, …, 1_ω) — would simply have irrational anti-diagonal (with 1_ω if diagonal is rational (without or with 0_ω), or vice versa on “ratiobal” and “irrational”;

* a truly complete enumeration or row-listing of all the fractional algebraic (inclined measurements) real numbers would have transcendental (curved, compounded continuously measurements) real diagonal and anti-diagonal numbers — and vice versa on “algebraic” and “transcendental”;

* a truly complete row-listing of all the fractional real numbers (complex numbers with no imaginary part — both algebraic and transcendental) would have complex (with real and imaginary part) diagonal and anti-diagonal numbers;

* a truly complete enumeration or row-listing of all the fractional complex numbers would have quaternion diagonal and anti-diagonal numbers;

* etc.

Even Ludwig Wittgenstein missed my point here -- you might want to read “An Editor Recalls Some Hopeless Papers” by Wilfrid Hodges. The very simple flaw in Hodges’ presented variant of Cantor’s anti-diagonal argument which claims to prove the “uncountability” of all the real numbers by “demonstrating” that there could not be any 1-to-1 correspondence between all the natural numbers and all the fractional real numbers is the common false belief that the arbitrary decimal place-value positional numeral system representation 0.d₁d₂d₃…d_n…, where d_n is in {0,1,2,3,4,5,6,7,8,9} for every natural number n, denotes the fractional expansion of either a rational (nonterminating, periodic) or an irrational (nonterminating, nonperiodic) real number between 0 and 1 when indeed just the rational numbers (since they have no period-length limit — in other words, any finite sequence of digits is a possible period) already exhaust them (on the other hand, the irrational numbers require the representation 0.d₁d₂d₃…d_n…1_ω due to the 1-1 correspondence, say π – 3 = 0. 0010010000111... in binary system, <0.0, 0.00, 0.001, 0.0010, 0.00100, 0.001001, …, π - 3> <--> <0, 0, 1, 0, 0, 1, …, 1_ω. [BenCawaling@Yahoo.com] BenCawaling (talk) 06:25, 26 March 2008 (UTC)

"Cantor's first proof" is new to me, and I have to say it's delightful. I agree that mathematicians generally believe the diagonal argument to be Cantor's first. However, I'm not completely convinced that this isn't really a diagonal argument in disguise. I need to think about this a bit. Dmharvey Talk 22:45, 6 Jun 2005 (UTC)

From the constructionists point view Cantors proof is "flawed". Let me first say that that I am not a constructionist, and that I do accept Cantor proof in full. Constructionists don't even consider it proven that there is no greatest number, and counterarguments won't bite. Here is why: Suppose there is a greatest natural number and call it Max. Then a kid in kindergarten can show that there is a number SuperMax = Max + 1 that is greater than Max. Contradiction right? Wrong, says the constructionist, because you haven't shown Max to exist, and therefore you cannot use it in any way in a proof of anything. Reductio ad absurdum arguments aren't generally accepted. Constructionists are "more rigorous" than most other mathematicians, probably in a sense that could be quantified exactly if the subset of the aximoms of ZFC that they accept is specified. The Axiom of Infinity is obviously not one of them. By the same token they can "proove fewer theorems" than most other mathematicians. The uncountability of the reals is obviously one of them. YohanN7 (talk) 02:20, 30 November 2007 (UTC)

Simplifications

I'm hesitant to make changes to the statement of the theorem and its proof, since we may want to retain historical accuracy, and I don't have access to Cantor's formulation. There's five things I would change:

1. R needs to be non-empty which is currently not required. (but see 2 below)
2. if we require that R have at least two points, then we can get rid of the endpointlessness requirement. (The first step of the proof would then read "pick the first element of the sequence that's not the largest element of R"; everything else stays the same.)
3. The proof emphasizes "greater than the one considered in the previous step" twice, but that's not needed; just always pick the first member of the sequence that works.
4. The proof should make a bit more explicit how the existence of c follows from the given properties of R. I.e.: How the sets A and B are defined.
5. The proof should make explicit how the density property is used.

What do people think about these changes? AxelBoldt 17:49, 30 March 2006 (UTC)

"contrary to what most mathematicians believe"

Those six words add nothing to the article but an unverified claim based on the personal experience of wikipedia editors. They're unencyclopedic language and not really appropriate. Night Gyr 22:15, 29 April 2006 (UTC)

As mentioned above, it's true that the claim hasn't been formally verified, though anecdotal evidence is strong. I don't see what's wrong with the language. How does removing this statement help our readers? AxelBoldt 00:20, 30 April 2006 (UTC)

Because it makes us less reliable as an encyclopedia to include information that can't be relied upon to be anything but the anecdotal experiences of our editors. It's the whole reason behind WP:V. Night Gyr 01:13, 30 April 2006 (UTC)

So if I change "most" to "many" and find two sources where mathematicians make the false claim, would you be happy? AxelBoldt 14:17, 30 April 2006 (UTC)

I still don't understand why wikipedia needs to state that a misconception exists more prominently than the truth itself. We're here to provide facts, and the fact that a large number mathematicians of mathematicians have a fact incorrect seems less important to the first phrase of the entire article than the fact itself. Night Gyr 02:26, 1 May 2006 (UTC)

I agree, it doesn't need to be in the first sentence, or even the first paragraph. But it should be documented nevertheless. AxelBoldt 14:42, 3 May 2006 (UTC)

The rationals

Copied from the main article:

(Could someone who understands explain why the set of rational numbers does not have property 4?)

Property 4 says that if you partition the set into two halves, then there must be a boundary point in the set. This is not true for the rationals: take as A the set of all rationals smaller than √2 and as B the set of all rational above √2. Then all rationals are covered, since √2 is irrational, so this is a valid partition. There is no boundary point in the set of rational numbers that separates A from B however. AxelBoldt 02:09, 23 May 2006 (UTC)

• Ok, but could you clarify a little please... in as much as if you have your boundry, and A contains the elements less that that boundry, and B the elements greater than it, the the boundry is not in A or B. Probably missing something here, just can't see what.
  • That isn't a partition. If c is in R, then for {A,B} to be a partition of R, c needs to be in A or in B. Eg, for property 4, c would have to be either the largest member of A or the smallest member B. Aij (talk) 02:13, 15 April 2008 (UTC)

@AxelBoldt - In your example, why can't the boundary point be the largest rational in A or the smallest rational in B? Then, all rationals less than the boundary will be in A and all rationals greater than the boundary will be in B. Or am I misunderstanding the meaning of the word "every point" as "every point in R"? Vijay (talk) 08:36, 11 January 2010 (UTC)

Axel may not be watching anymore — he made that comment in 2006.

Anyway, for A and B as defined, A doesn't have a largest element, and B doesn't have a smallest element. --Trovatore (talk) 09:40, 11 January 2010 (UTC)

Complete is the wrong word

Technically, the real numbers are not complete. Of course, the extended reals are complete. It may be better to remove the completeness requirement and leave the "i.e.". Really, all we need is what Rudin calls the "least-upper-bound property." That is, the least upper bound of any set that is bounded from above exists (and the similar claim about sets bounded from below and infimum). --TedPavlic 15:32, 7 March 2007 (UTC)

• Additionally, the partition definition only makes sense if the element c is greater than or EQUAL to every element in A and less than or EQUAL TO every element in B. If c is not included in A or B, then (A,B) cannot be a partition of R. --TedPavlic 21:04, 7 March 2007 (UTC)
  • The text seems fine, it does not say that c is greater than every element in A and less than every element in B.--Patrick 11:23, 7 June 2007 (UTC)
    • I you don't allow for "or equal to" in at least one place i'ts impossible to find such a partition. (Present a counterexample please in case I'm wrong.)YohanN7 (talk) 00:40, 30 November 2007 (UTC)
    • I correct myself. It is perfectly correct as it stands. Still, I obviously wasn't the only one that fell into the trap. If a wording with "... or equal to" is equivalent to the current wording that would be preferable.YohanN7 (talk) 00:51, 30 November 2007 (UTC)
• Note: the term "gapless" may be better than "complete". --TedPavlic 21:17, 7 March 2007 (UTC)

Complete is the correct word. From the article you linked: "In order theory and related fields such as lattice and domain theory, completeness generally refers to the existence of certain suprema or infima of some partially ordered set. Notable special usages of the term include the concepts of complete Boolean algebra, complete lattice, and complete partial order (cpo). Furthermore, an ordered field is complete if every non-empty subset of it that has an upper bound within the field has a least upper bound within the field, which should be compared to the (slightly different) order-theoretical notion of bounded completeness. Up to isomorphism there is only one complete ordered field: the field of real numbers (but note that this complete ordered field, which is also a lattice, is not a complete lattice)." Ossi 19:45, 8 September 2007 (UTC)

The article linked to complete lattices, and you're right the reals are not a complete lattice. They are a complete metric space, but that is not the subject here. I reworded it a little. I don't think "gapless" is better - we should try to explain things, but not to the point of inventing our own terminology. — Carl (CBM · talk) 13:15, 23 July 2007 (UTC)

I added an External Link from this page to Theorem of the Day, no. 19. It seemed an innocent enough thing to do. Theorem of the Day is an academic project which tries to bring beautiful mathematics to a wider public without sacrificing rigour. It has over 100 theorems on offer now and is reasonably widely known. I am new to Wikipedia, however, so if this is what is regarded as spam, then maybe I rethink how I publicise my project? Charleswallingford (talk) 22:56, 24 June 2008 (UTC)

Cantor Anti-Diagonal Argument — Clarifying Determinateness and Consistency in Knowledgeful Mathematical Discourse

Perhaps my unfinished manuscript "Cantor Anti-Diagonal Argument -- Clarifying Determinateness and Consistency in Knowledgeful Mathematical Discourse" would be useful now to those interested in understanding Cantor anti-diagonal argument. I was hoping to submit it to the Bulletin of Symbolic Logic this year. Unfortunately, since 1 January 2008, I have been suffering from recurring extremely blurred vision due to frequent “exploding optical nerves” brought on by my diabetes (I can’t afford laser eye surgery) and I had only about 20 productive days in the last 8 months. At this rate, it would take me a long while to finish my paper or may not be able to complete it if I go permanently blind soon. I just hope my endeavors to clarify mathematical infinity and modern logic would reach the next (if not the present) generations of mathematicians, philosophers, and logicians. [BenCawaling@Yahoo.com] BenCawaling (talk) 08:08, 4 September 2008 (UTC)

Proposed Changes to Article

I am very happy to see a Wikipedia article about Cantor's first uncountability proof. Since I have studied Cantor's 1874 article and some of his correspondence, I started adding material and making some changes. The result of this work can be found at: Talk:Cantor's first uncountability proof/Temp. I hope you find my revisions interesting and relevant. I'm looking forward to your suggestions, modifications, and feedback. Here's a section-by-section summary of my revisions:

Introduction: Made some changes and mentioned two controversies that have developed around Cantor's article. The "emphasis" controversy ("Why does Cantor's article emphasize the countability of the set of real algebraic numbers?") is already discussed in the current article. The "constructive/non-constructive" controversy concerns Cantor's proof of the existence of transcendental numbers.

Development and Publication: Expanded the current "Publication" section by adding material that comes mostly from Cantor's correspondence. Like the current section, this new section discusses the "emphasis" controversy, but I did add some material here.

The Article: Replaces the current "The theorem" section. Contains statements of the theorems that Cantor proves in his article. Also, uses Cantor's description of his article to bring out the article's structure. This structure is the key to handling the "constructive/non-constructive" controversy.

The Proofs: Contains proofs of Cantor's theorems.

Cantor’s Method of Constructing Transcendental Numbers: Replaces the current "Real algebraic numbers and real transcendental numbers" section. Also, discusses the "constructive/non-constructive" controversy.

I have also added a "Notes" section, and I have added references to the current "References" section.

I highly recommend reading Cantor's original article, which is at: "Über eine Eigenschaft des Ingebriffes aller reelen algebraischen Zahlen". A French translation (which was reviewed and corrected by Cantor) is at: "Sur une propriété du système de tous les nombres algébriques réels". Unfortunately, I have not found an English translation on-line. However, an English translation is in: Volume 2 of Ewald's From Immanuel Kant to David Hilbert: A Source Book in the Foundations of Mathematics (ISBN 9780198532712).

Most of the material I added to this Wikipedia article comes from Cantor's article, Cantor's correspondence, Dauben's biography of Cantor (ISBN 0674348710), and the article "Georg Cantor and Transcendental Numbers".

Finally, I wish to thank all the people who have worked on this Wikipedia article. Without the excellent structuring of your article and the topics you chose to cover, I suspect that I would not have written anything. (This is the first time I've written for Wikipedia.) It's much easier to add and revise rather than develop from scratch. RJGray (talk) 23:30, 5 May 2009 (UTC)

Rewrote the section "Cantor’s method of constructing transcendental numbers" and renamed it "Is Cantor’s proof of the existence of transcendentals constructive or non-constructive?" The old section did not explain this constructive/non-constructive controversy. The new section quotes mathematicians on different sides of the controversy, analyzes their versions of "Cantor's proof," looks at relevant letters of Cantor's, mentions some computer programs, and then shows Cantor's diagonal method in a simpler context -- namely, generating the digits of an irrational (rather than the more difficult job of generating the digits of a transcendental).

--RJGray (talk) 02:55, 5 August 2009 (UTC)

Oops, I forgot to thank Michael Hardy for the feedback that he has given me on my proposed changes. His feedback made me realize that my old section was inadequate. I hope that my new section is more adequate -- I welcome your feedback on it. --RJGray (talk) 03:11, 5 August 2009 (UTC)

I suggest scrapping the article.

The following discussion is closed. Please do not modify it. Subsequent comments should be made in a new section.

The four properties say nothing about countability. One could reword the entire theorem to state:

A set G composed of linearly ordered sets A and B with all members of A < B contains a member c that is indeterminate.

This is just another flavour of Dedekind's definition of real number. Defining a real number does not make it 'countable', whatever countable means for indeed it is not defined in any way. What is being defined first here - countability or a real number? The Diagonal Argument is very different to this inconsequential theorem. Both the theorem and diagonal argument are flawed. 91.105.179.213 (talk) 10:21, 19 January 2010 (UTC)

You're confused. The set G you refer to DOES NOT generally contain a member between A and B; in particular, if G is countable then it can be partitioned into such sets A and B such that G contains no such element. And you haven't defined "indeterminate". Countability is used in order to reach the conclusion. If you missed that, you haven't read carefully. Michael Hardy (talk) 14:29, 19 January 2010 (UTC)

OK, more long windedly:

• "A set G composed of linearly ordered sets A and B with all members of A < B"

No, that's not right. What is written about is a linearly ordered set G. If is not enough for A and B separately to be linearly ordered; the whole set G is supposed linearly ordered.

• "with all members of A < B"

I presume this means every member of A is less than every member of B; it should have said so.

• "contains a member c that is indeterminate"

What does "indeterminate" mean? The article claims

• c is not a member of either A or B, and
• every member of A is less than c, and
• c is less than every member of B.

Moreover, the article says c is not a member of G! If c is not a member of G, then G does not "contain" c! Yet, you've set "contains a member that is...".

Is it true that if a linearly ordered set G is partitioned into subsets A and B with every member of A less than every member of B, then such an element c exists? No. It is not true. For example, let G be the set of all real numbers, let A be the set of all negative numbers, and let B be the set of all non-negative numbers, i.e. B contains all positive numbers and 0. Does G contain any member c that is greater than every member of A and less than every member of B? No, it does not.

The element c is explicitly stated not to be a member of G, yet you say G "contains a member that..." etc.

Countability was used in reaching the conclusion. The conclusion was not that the countable set G contains the number c, but rather that the set of real numbers contains the number c, and G does not contain the number c.

You need to pay attention. Michael Hardy (talk) 14:45, 19 January 2010 (UTC)

Is it I or you who needs to pay attention? The article states:

Property 4 has the following least upper bound property: For any partition of R into two nonempty sets A and B such that every member of A is less than every member of B, there is a boundary point c (in R), so that every point less than c is in A and every point greater than c is in B.

Just replace R with G and you have exactly what I claimed:

A set R composed of linearly ordered sets A and B with all members of A < B contains a member c that is indeterminate.

Okay. So why is c indeterminate? Because your article states it is so: c is a boundary point in R.

Mr. Hardy, I object to your patronizing tone. Do not talk down to me! 91.105.179.213 (talk) 20:57, 19 January 2010 (UTC)

Please refrain from this sort of argument on article talk pages. See WP:TALK. 91, if you want to discuss this sort of issue with Dr Hardy, I suggest you contact him directly. --Trovatore (talk) 21:05, 19 January 2010 (UTC)

Excuse me? And who might you be? I don't give a hoot whether he is a Mr. or a Dr or whatever. He better not be talking down to me! Trovatore - I suggest you pay attention! Before you start blurting out baseless accusations. Look Travo, just stay out of this conversation. Hardy does not need your help. I certainly have not violated any WP:TALK rules. You should be directing your comments at him! Besides, this is the discussion page - outside of this page, I have nothing to say to Hardy. 91.105.179.213 (talk) 22:06, 19 January 2010 (UTC)

Actually you have gone against talk page guidelines — talk pages are not for general discussion of the subject matter. --Trovatore (talk) 01:27, 20 January 2010 (UTC)

"91.105.179.213", I do think you need to pay closer attention. You wrote about

"A set G composed of linearly ordered sets A and B"

The fact is, the least upper bound property was attributed here to the linearly ordered set of real numbers, not to just any linearly ordered set. And the proof shows that any countable subset lacks that property. Essential use of countability was made in the argument. Therefore this does indeed have to do with countability. Michael Hardy (talk) 00:21, 20 January 2010 (UTC)

....OK, a bit more long-windedly again:

• There are some linearly ordered sets that when partitioned into subsets every member of one of which is less than every member of the other, contain a boundary point (which must of course be a member of one of those two subsets or the other). The set of real numbers in their usual order is such a case.
• There are some linearly ordered sets G that can be partitioned into subsets every member of one of which is less than every member of the other such that there is no such boundary point within G. Such a boundary point will be found within a completion of G, of which G itself is a subset. The set of all rational numbers is an example.

Your statement about an "indeterminate point" seems to say that that indeterminate point must be within G. But the example of the rational numbers shows that that is not true.

At any rate, it's clear that you haven't understood the argument. You also haven't said what you mean by "indeterminate". Michael Hardy (talk) 00:40, 20 January 2010 (UTC)

That I haven't understood the argument is correct. However, I think the problem is not with my understanding but rather with the argument. The article states 4 properties that are not equivalent to the same requirement for rational numbers to be considered countable. In other words, there is nothing that demonstrates on any level that the real numbers can be placed into a one to one correspondence with the natural numbers or not. Well, if this is what was written in Cantor's original work, it does not matter whether it is correct or not as far as Wikipedia is concerned. After all, the goal is to report the argument, not to debate it. Try not to be so arrogant Hardy. 91.105.179.213 (talk) 05:23, 20 January 2010 (UTC)

The four requirements have nothing to do with the rationals being countable. The four requirements are satisfied by the real numbers in their usual order, not the rational numbers. The claim is that any linearly ordered set satisfying those four requirements is uncountable. The proof shows that any countable subset must fail to contain every member of the linearly ordered set. The way it is proved is by showing that regardless of which countable set you pick, there will be some elements of the initially considered linearly ordered set that fail to be contained within it.

If I try to explain the argument to you, does that make me "arrogant"? Michael Hardy (talk) 08:01, 20 January 2010 (UTC)

Talking down to someone by saying that person needs to pay attention is disdainful, arrogant and unnecessary.

The four requirements have nothing to do with the rational numbers being countable – Yes, that would be correct. As I stated, the 4 properties are Dedekind's definition of a real number. So, what definition of countability does Cantor assume in his definition (Then R is not countable) that the real numbers are uncountable? A few circular references here do not quite add up. Do you know where one can find the original text on the internet? The wiki link to this text is broken - Geocities no longer exists. I read German and am curious to see exactly how Cantor phrased it. 91.105.179.213 (talk) 10:00, 20 January 2010 (UTC)

The four properties are consequences of Dedekind's definition. Those four properties do not apply only to real numbers, but also to some other linearly ordered sets.

The definition of "countable" is just that a countable set is one whose members are in a sequence: a first one, a second one, a third one, etc., so that each is reached after only finitely many steps. I don't know what you mean by the word "definition"? You say "his definition (Then R is not countable)". That's not a definition; that's a conclusion. And you write of "his definition[...]that the real numbers are uncountable". Again: that's not a definition; that's a conclusion. Using words in odd ways like this that are different from what everybody else uses does not help people understand what you're saying. Michael Hardy (talk) 17:59, 20 January 2010 (UTC)

The four properties are Dedekind's definition of real number. These are not consequences of as you claim. However this is immaterial to my understanding. The problem lies with the formulation of the theorem itself.

Let me see if I can explain this more simply:

Article states: Suppose a set R has the following 4 properties then R is not countable.

This logic is equivalent to stating that proposition P implies NOT proposition C. P can be the statement of the 4 properties. But what is C? Countability, one would imagine. If so, where is countability defined? If this Cantor's first countability proof, then where did he define countability seeing the proof uses the word countable as in R is not countable. Make sense to you now? 91.105.179.213 (talk) 18:39, 20 January 2010 (UTC)

Dedekind's definition of real number identified a real number with a Dedekind cut in the set of rational numbers. There are other linearly ordered sets than the rationals with which one can do the same thing, and get different linearly ordered sets satisfying the same properties, but those are not the set of real numbers.

I believe the way Cantor wrote the argument originally is simply that he set no sequence contains all real numbers, understanding sequence to mean something that has a first element, a second, a third, and so on, each element being reached after some finite number of steps. He also wrote in the same paper that there is a sequence containing all real algebraic numbers, and showed how to construct it.

I looked for the paper in pdf format and was surprised that I couldn't find it quickly. Michael Hardy (talk) 06:01, 21 January 2010 (UTC)

I can't comment about Cantor's sequence containing all real algebraic numbers. My first reaction is it does not sound correct. Regarding your comments on linearly ordered sets, again, property 4 of the theorem makes it clear Cantor is dealing with real numbers (least upper bound property implies c is real number).

You could not find the paper quickly - does this imply you found it? If so, please provide a link to it. 91.105.179.213 (talk) 07:42, 21 January 2010 (UTC)

I have not found it on the web. Michael Hardy (talk) 21:30, 21 January 2010 (UTC)

Saying that the four conditions mentioned in the article are Dedekind's definition of "real number" seems quite confused. Dedekind identified a real number with Dedekind cut within the linearly ordered set of rational numbers. Such a cut is a partition of the rationals into two sets A, B, with every element of A less than every element of B. It is not the case that the rationals contain a boundary point between those two sets, which would have to be a member of one or the other of the two sets. If that were true, what Dedekind was doing wouldn't have worked. It is true that if you partition the real numbers in that way, then the set of reals contains such a boundary point. But that is certainly not Dedekind's definition of the reals. Nor are the reals the only linearly ordered set with that property; there are others, essentially different from the reals in the sense of not being isomorphic as an ordered set. Hence the four conditions fall short of fully characterizing the reals as a linearly ordered set. Stating the four conditions is simpler than what Dedekind did. Michael Hardy (talk) 21:38, 21 January 2010 (UTC)

Read property 3 very carefully. Can you show me any other linearly ordered set (besides the rational numbers) of which it can be said: between any two members there is always another? Allow me to answer this question. NO. Property 3 cannot be 'digested' - such an assumption requires a huge leap of faith. So, since no other linearly ordered set meets this requirement, the matter is settled - property 3 can be reworded as "Assume R contains rational numbers". Property 4 becomes a bygone conclusion and Cantor's proof is just another flavour of Dedekind's ideas. 91.105.179.213 (talk) 22:03, 21 January 2010 (UTC)

In my previous post I was assuming the real numbers are not well-defined. Yes, yes. I know you think the real numbers are well-defined. So I too will assume they are well-defined. Even so, I still see many problems with the theorem.

1. Stating the 4 properties, says that R contains real numbers.

2. The theorem states that if R meets these conditions, then R is not countable without defining countable. If one accepts that one can first define not countable and then determine countable as its compliment, then it might appear that Cantor's first ideas had nothing to do with the notion of a one-to-one correspondence of a subset with its proper subset as the definition of countable.

3. The theorem could be stated as follows: Any set that has at least these 4 properties is not countable. But to state it this way, is the same as saying the real numbers are not countable by virtue of their design/architecture.

4. In set theory, if a proper subset contains a certain property, then it implies the superset must also posses that property by inclusion of the subset. So if the rational numbers are countable, this would imply the real numbers must therefore also be countable. 91.105.179.213 (talk) 05:36, 22 January 2010 (UTC)

Sorry, you're quite wrong. You wrote:

Can you show me any other linearly ordered set (besides the rational numbers) of which it can be said: between any two members there is always another? Allow me to answer this question. NO.

That is nonsense. I can easily show you many such sets. One of them is the real numbers. Another is the real algebraic numbers. Another is rational functions in one variable with real coefficients with a certain natural ordering. Michael Hardy (talk) 05:42, 22 January 2010 (UTC)

You wrote:

In set theory, if a proper subset contains a certain property, then it implies the superset must also posses that property by inclusion of the subset.

That is nonsense. The set {1, 2, 3} has the property of finiteness. The set {1, 2, 3, ...} does not. Michael Hardy (talk) 05:47, 22 January 2010 (UTC)

You wrote:

property 3 can be reworded as "Assume R contains rational numbers". Property 4 becomes a bygone conclusion

That is nonsense. The set of all real algrebraic numbers contains the set of all rational numbers, but does not satisfy property 4 at all. Michael Hardy (talk) 05:51, 22 January 2010 (UTC)

It's quite amazing to me how you picked out what you felt like answering and ignored the big elephant in the room. So, I respond as follows: I posted a second comment that clarified my first assumption. Your example of a finite and infinite set is irrelevant - all the sets we are discussing are infinite: naturals, rationals and reals. So how about addressing the real issues -

1. Stating the 4 properties, says that R contains real numbers.

2. The theorem states that if R meets these conditions, then R is not countable without defining countable. If one accepts that one can first define not countable and then determine countable as its compliment, then it might appear that Cantor's first ideas had nothing to do with the notion of a one-to-one correspondence of a subset with its proper subset as the definition of countable.

3. The theorem could be stated as follows: Any set that has at least these 4 properties is not countable. But to state it this way, is the same as saying the real numbers are not countable by virtue of their design/architecture.

4. In set theory, if a proper subset contains a certain property, then it implies the superset must also posses that property by inclusion of the subset. So if the rational numbers are countable, this would imply the real numbers must therefore also be countable.

Or is it that you simply can't bear to admit Cantor's theorem and his theories in general are cranky?

Hardy, I really don't want you becoming upset over this. Just take a deep breath and stay calm. Try to be your normal, cool-headed, objective self.

91.105.179.213 (talk) 09:14, 22 January 2010 (UTC)

According to WP:TALK: "Do not use the talk page as a forum or soapbox for discussing the topic. The talk page is for discussing improving the article." We usually allow a little freedom to discuss things if it appears that it might lead to article editing, but in this case I think the discussion has become tangential and unlikely to lead to improvements to the article. Therefore, I would suggest ending it here.

Concrete suggestions about how to modify the article are welcome; please use a new section for them. My recommendation for 91.105.179.213 is to consult a good introductory book on set theory. Any such book will have a clear explanation of countability. — Carl (CBM · talk) 12:58, 22 January 2010 (UTC)

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

My advice to Carl

Um, no thanks. I really don't need your advice. I was prepared to stop editing but Hardy (your math-god) decided he wanted to pursue the topic. So, I suggest you run off to your leader and tell him to consult a good book on mathematical logic. While you are at it, get one yourself.

Hardy: Seems to me every time you don't know what you are talking about, you order one of your drones to terminate the conversation? Frankly I don't care to pursuit it.

What an arrogant, foolish lot you are. 91.105.179.213 (talk) 14:01, 22 January 2010 (UTC)